Q.
Two magnets of equal mass are joined at 90${^\circ}$ each other as shown in figure. Magnet $N_{1}\ S_{1}$ has a magnetic moment $\sqrt{3}$ times that of $N_{2}\ S_{2}$. The arrangement is pivoted so that it is free to rotate in horizontal plane. When in equilibrium, what angle should $N_{1}\ S_{1}$ make with magnetic meridian?
A) 75${^\circ}$
B) 60${^\circ}$
C) 30${^\circ}$
D) 45${^\circ}$
Explanation
(c) In equilibrium, the resultant magnetic moment will be along magnetic meridian. Let $N_{1}S_{1}$ make $\angle\theta$ with resultant
$tan\theta = \frac{M_{2}}{M_{1}} = \frac{M}{\sqrt{3}M} = \frac{1}{\sqrt{3}}\ \therefore\theta = 30{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.